fmn and tfmn in the 6/9 Magic-Square Problem

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1. From square patterns to the tfmn problem

The general theory of square-valued patterns first classifies the 4/9 and 5/9 configurations and translates squarehood of the selected entries into systems of quadrics. The next level is 6/9. Among its positional types are patterns formed by two disjoint arithmetic progressions in the same direction. Up to rotations and reflections there are two such types, represented by

\underbrace{AEJ}_{\mathrm{AP}}\sqcup \underbrace{BFG}_{\mathrm{AP}}=ABEFGJ, \qquad \underbrace{AFH}_{\mathrm{AP}}\sqcup \underbrace{BDJ}_{\mathrm{AP}}=ABDFHJ.

In the general form of a magic square, the two progressions in each pair are parallel and have the same oriented difference. If the first progression is parametrized by (m,n) with scale u and the second by (p,q) with scale v, their differences have the form

\Delta_1=4u^2f(m,n),\qquad \Delta_2=4v^2f(p,q).

The third independent quadric of each of these two patterns states exactly that their oriented differences agree. Hence, after parametrizing the red conics, the entire residual system is the single equation

u^2f(m,n)=v^2f(p,q).

It is solvable in nonzero rational scales u and v exactly when f(m,n) and f(p,q) lie in the same square class. The function tf is introduced precisely as a canonical coordinate of this class. For ABEFGJ and ABDFHJ the condition is not merely necessary but sufficient: after matching the differences, the two progression centers directly recover the unique triple E,x,y.

Preceding theory of 6/9 patterns Complete derivation of the two parallel patterns

2. Area and progression difference

For nonzero rational m,n, define

f(m,n)=mn(m^2-n^2) =mn(m-n)(m+n).

This form arises from the classical parametrization of Pythagorean triples:

(m^2-n^2)^2+(2mn)^2=(m^2+n^2)^2,\qquad \frac{(m^2-n^2)(2mn)}2=f(m,n).

For m>n>0, the value f(m,n) is the area of the right triangle with legs m²−n² and 2mn. For arbitrary orders and signs of the parameters, f also records orientation.

The connection with arithmetic progressions of squares follows from the identities

\begin{aligned} (m^2+n^2)^2-4f(m,n)&=(-m^2+2mn+n^2)^2,\\ (m^2+n^2)^2+4f(m,n)&=(m^2+2mn-n^2)^2. \end{aligned}

Consequently, the three expressions

V-4f(m,n),\qquad V,\qquad V+4f(m,n), \qquad V=(m^2+n^2)^2,

are squares and form an arithmetic progression with oriented difference 4f(m,n). This identity underlies the application of the fmn construction to the eight progressions in a magic square.

3. The squarefree part

For a rational number r∈ℚ×, let t(r) denote the squarefree part of r. It is the unique squarefree integer with the same sign as r for which

r=t(r)q^2,\qquad q\in\mathbb Q_{>0}.

Write the rational number in terms of its p-adic valuations:

r=\operatorname{sgn}(r)\prod_p p^{v_p(r)}, \qquad v_p(r)\in\mathbb Z,

then

t(r)=\operatorname{sgn}(r) \prod_{v_p(r)\ \mathrm{odd}}p.

Parity has its usual meaning for negative valuations as well. Removing the residue of every vₚ(r) modulo 2 leaves only even exponents, and their product is a rational square. Uniqueness follows because the sign of t(r) is fixed by the sign of r, while every prime exponent in a squarefree representative can only be 0 or 1 and must agree with vₚ(r) modulo 2.

For any representation r=a/b with nonzero integers a,b, this gives the particularly simple formula

t\!\left(\frac ab\right)=t(ab),

because a/b=ab/b². The formula is independent of the chosen numerator and denominator: if a/b=c/d, then a/c=b/d, and hence ab/(cd)=(a/c)² is a rational square.

In the language of square classes, this means

t(A)=t(B) \iff \frac AB\in(\mathbb Q^\times)^2, \qquad A,B\in\mathbb Q^\times.

Thus t(r) is the canonical integral representative of the class of r in ℚ×/(ℚ×)².

Definition of the function tf

\operatorname{tf}(m,n)=t\!\left(f(m,n)\right), \qquad m,n\in\mathbb Q,\quad f(m,n)\ne0.

Thus tf(m,n) is the squarefree part of f(m,n). When f(m,n)=0, the progression is constant and tf is undefined.

4. tf(m,n) as a congruent number

Let m>n>0 be integers and write

f(m,n)=Tq^2,\qquad T=\operatorname{tf}(m,n)>0.

Divide the sides of the corresponding Pythagorean triangle by q:

A=\frac{m^2-n^2}{q},\qquad B=\frac{2mn}{q},\qquad H=\frac{m^2+n^2}{q}.

This gives a rational right triangle of area

\frac{AB}{2}=\frac{mn(m^2-n^2)}{q^2}=T.

Therefore a positive value of tf(m,n) is the squarefree part of the area and is also a congruent number. Distinct pairs (m,n) may have the same squarefree part and hence determine the same T.

5. Criteria for equality of tf values

Theorem

Let Fᵢ=f(mᵢ,nᵢ) be two nonzero integral values of f and let G=gcd(|F₁|,|F₂|). The following conditions are equivalent:

$t(F_1)=t(F_2)$ ;
$F_1/F_2$ is a square in ℚ;
$F_1F_2$ is a positive perfect square;
F₁,F₂ have the same sign, and $|F_1|/G$ and $|F_2|/G$ are perfect squares.

Proof

Write Fᵢ=Tᵢqᵢ², where Tᵢ=t(Fᵢ) is the squarefree part of Fᵢ. The quotient F₁/F₂ is a square exactly when the sign and the parity of every prime exponent agree, that is, exactly when T₁=T₂. The product F₁F₂ is a positive square under the same condition.

If T₁=T₂=T, then G=|T|gcd(q₁,q₂)², so both quotients |Fᵢ|/G are squares. Conversely, if those two quotients are squares and the signs agree, their quotient F₁/F₂ is a rational square. Hence all four conditions are equivalent.

6. Matching the scales of progressions

Multiplying all three terms of a progression of squares by λ² preserves squarehood and multiplies its difference by λ². Thus two parametrized progressions with values F₁=f(m₁,n₁) and F₂=f(m₂,n₂) admit a common oriented difference exactly when there exist nonzero α,β∈ℚ such that

4\alpha^2F_1=4\beta^2F_2.

After cancellation, this equality is precisely the second condition of the preceding theorem. Therefore,

\operatorname{tf}(m_1,n_1)=\operatorname{tf}(m_2,n_2) \iff \text{the progressions scale to a common oriented difference}.

Constructively, if F₁=Tu² and F₂=Tv², multiply the first progression by v² and the second by u². Their differences both become 4Tu²v². This is the arithmetic content of equality between tf values.

7. Parameter symmetries

Some equalities between tf values arise from parameter substitutions that preserve the square class or merely reverse the orientation of the progression:

Substitution	Effect on f	Meaning
$(m,n)\mapsto(\lambda m,\lambda n)$	$f\mapsto\lambda^4f$	common parameter scale
$(m,n)\mapsto(-m,-n)$	$f\mapsto f$	simultaneous sign change
$(m,n)\mapsto(n,m)$	$f\mapsto-f$	reversal of the progression
$(m,n)\mapsto(m,-n)$	$f\mapsto-f$	reversal of the progression
$(m,n)\mapsto(m+n,m-n)$	$f\mapsto4f$	the same normalized progression

The last row follows immediately from the factorization f(a,b)=ab(a+b)(a−b):

\begin{aligned} f(m+n,m-n) &=(m+n)(m-n)\cdot2m\cdot2n\\ &=4f(m,n). \end{aligned}

The middle square V=(m²+n²)² is also multiplied by 4 under the same substitution. Hence the ratio 4f/V, namely dir, is unchanged. Since the factor 4 is a square, the value of tf is preserved as well. When unoriented progressions are classified, opposite signs of tf are identified as well.

8. The self-recurrence of tf

A progression of squares itself supplies a new parameter pair. Set

V=(m^2+n^2)^2,\qquad D=4f(m,n).

If r=−m²+2mn+n², s=m²+n², and w=m²+2mn−n², then V=s², V−D=r², and V+D=w². Therefore

\begin{aligned} f(V,D) &=VD(V^2-D^2)\\ &=s^2\cdot4f(m,n)\cdot r^2w^2\\ &=f(m,n)(2srw)^2. \end{aligned}

The quotient f(V,D)/f(m,n) is a square, and the square-class theorem gives the identity

\operatorname{tf}(m,n) = \operatorname{tf}\!\left((m^2+n^2)^2,\;4f(m,n)\right).

9. From fixed tf to F7+

A fixed positive value T=tf(m,n) determines the congruent-number elliptic curve

E_T:\quad y^2=x^3-T^2x.

This is not merely a source of examples. After quotienting parameter pairs by their common scale, there is an exact bijection between all nondegenerate rational pairs of the fixed square class and the nontrivial rational points of E_T, up to the sign of y. The separate F7+ article derives both maps, proves that they are inverse, and describes the exact degenerate boundary.

F7+: the complete bijection →

10. What remains of the 6/9 problem

For the two parallel 6/9 classes, equality of tf values solves the entire system of selected square entries. In ABEFGJ the two progression centers occupy E and G; in ABDFHJ they occupy A and J. Once the common difference is chosen, those centers give E,x,y by linear formulas, so no further algebraic placement conditions remain.

The limitation lies elsewhere. Equality of tf does not guarantee positivity of all nine entries, pairwise distinctness, or the absence of extra squares in the three unselected positions. Moreover, in the other fourteen positional types the red progressions intersect or are absent, and their three quadrics no longer reduce to a single comparison of differences.

The common Kummer surface of the patterns →Generating equal tf values →The complete F7+ bijection The early classification